# Closed form for $\sum_{k=1}^\infty\frac{H_k^{(m)}}{k^n}$

Let's define

$$\sigma(m,n)=\sum_{k=1}^\infty\frac{H_k^{(m)}}{k^n}$$

where $$H_k^{(m)}=\sum_{n=1}^{k}\frac{1}{n^m}$$ is the k-th generalized harmonic number of order $$m$$.

In mathworld site Eq (20), I found

$$\sigma(m\text{ even},n\text{ odd})=\frac12\left[\binom{m+n}{m}+1\right]\zeta(m+n)+\zeta(m)\zeta(n)$$

$$-\sum_{j=1}^{m+n}\left[\binom{2j-2}{m-1}+\binom{2j-2}{n-1}\right]\zeta(2j-1)\zeta(m+n-2j+1)\label{1}\tag{1}$$

and $$\sigma(m\text{ odd},n\text{ even})=-\frac12\left[\binom{m+n}{m}+1\right]\zeta(m+n)$$

$$+\sum_{j=1}^{m+n}\left[\binom{2j-2}{m-1}+\binom{2j-2}{n-1}\right]\zeta(2j-1)\zeta(m+n-2j+1)\label{2}\tag{2}$$

I know that \eqref{2} follows from \eqref{1} by using the well-known indentity

$$\sum_{k=1}^\infty \frac{H_k^{(m)}}{k^n}+\sum_{k=1}^\infty \frac{H_k^{(n)}}{k^m}=\zeta(m)\zeta(n)+\zeta(m+n)$$

The proof of \eqref{1} may be found here but I am looking for different ones if possible. Thanks